In a right circular cone, the radius of its base is 7 cm and its height is 24 cm. A cross-section is made through the mid-point of the height parallel to the base. The volume of the upper portion is :

$$\eqalign{ & r = 7{\text{ cm, }}h{\text{ = 24 cm}} \cr & {\text{Now, }}\vartriangle {\text{AOB}} \sim \vartriangle {\text{COD}} \cr & {\text{So, }}\frac{{OA}}{{OC}} = \frac{{AB}}{{CD}} \cr & \Rightarrow \frac{h}{{\frac{h}{2}}} = \frac{r}{{CD}} \cr & \Rightarrow CD = \frac{r}{2} \cr} $$
∴ Volume of upper portion :
$$\eqalign{ & = \frac{1}{3}\pi {\left( {\frac{r}{2}} \right)^2}\left( {\frac{h}{2}} \right) \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12} \right){\text{ c}}{{\text{m}}^3} \cr & = 154{\text{ c}}{{\text{m}}^3} \cr} $$